There are two hyperbolic closed 3-manifolds, but I don't know whether they are homeomorphic or not. The only thing I know is that the Cayley graphs of their fundamental groups are quasi-isometric.
My question is:
Will the manifolds be homeomorphic?
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7$\begingroup$ If the manifolds are compact then their Cayley graphs are quasi-isometric to hyperbolic 3-space; so you get no info this way. $\endgroup$– Anton PetruninCommented Aug 17, 2013 at 1:35
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$\begingroup$ @Anto: Will you explain it ? $\endgroup$– yanqingCommented Aug 17, 2013 at 7:25
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1$\begingroup$ @Anton: Also, please consider posting your comment (and any further explanation) as an answer. Thank you. $\endgroup$– Ricardo AndradeCommented Aug 17, 2013 at 7:40
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5$\begingroup$ @RicardoAndrade I conjecture that Anton does not think that this is a research level discussion -- I know I think it more appropriate for stackexchange. $\endgroup$– Igor RivinCommented Aug 17, 2013 at 15:17
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1 Answer
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$\newcommand{\HH}{\mathbb{H}}$Here is an expansion of what Anton is saying.
Suppose that $M$ is a closed hyperbolic three-manifold. It follows that the universal cover of $M$ is $\HH^3$: hyperbolic space. The covering map of $M$ comes with a deck group - namely there is an action of $\pi_1(M)$ on $\HH^3$ so that the quotient $\pi_1(M) \backslash \HH^3$ is homeomorphic to $M$.
Now fix a set $S$ of generators for $\pi_1(M)$. Let $\Gamma = \Gamma(M, S)$ be the Cayley graph for $\pi_1(M)$ relative to $S$. Also, fix a point $x$ of $\HH^3$. Let $\rho_x{:}\,\Gamma \to \HH^3$ be the orbit map: namely $\rho_x(g) = g \cdot x$ for vertices of $\Gamma$, and all edges of $\Gamma$ are sent to geodesics. The Švarc-Milnor lemma says that $\rho_x$ is a quasi-isometry.
It follows that the Cayley graphs for any pair of closed hyperbolic three-manifolds are quasi-isometric. Thus homeomorphism types cannot be distinguished this way.
